Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{t^3 - 4t^2 - 5t}{-6t^3 - 6t^2 + 180t}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ x = \dfrac {t(t^2 - 4t - 5)} {-6t(t^2 + t - 30)} $ $ x = -\dfrac{t}{6t} \cdot \dfrac{t^2 - 4t - 5}{t^2 + t - 30} $ Simplify: $ x = - \dfrac{1}{6} \cdot \dfrac{t^2 - 4t - 5}{t^2 + t - 30}$ Since we are dividing by $t$ , we must remember that $t \neq 0$ Next factor the numerator and denominator. $ x = - \dfrac{1}{6} \cdot \dfrac{(t - 5)(t + 1)}{(t - 5)(t + 6)}$ Assuming $t \neq 5$ , we can cancel the $t - 5$ $ x = - \dfrac{1}{6} \cdot \dfrac{t + 1}{t + 6}$ Therefore: $ x = \dfrac{ -t - 1 }{ 6(t + 6)}$, $t \neq 5$, $t \neq 0$